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3 years ago

When patterning a shotgun, what is a sufficient percentage of pellets within a 30-inch circle?.

1 answer:

8 0

55% to 60%
The pattern of pellets within a 30-inch circle should be of a proper, even density to ensure a clean kill. The pattern should contain a sufficient percentage of the load, which should be at least 55% to 60%.

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*REVERSE PROPORTION*

muminat

9514 1404 393

Answer:

33.3%

Step-by-step explanation:

The selling price of £42 is (1 +40%) times the total purchase price.

1.40 × purchase price = £42

purchase price = £42/1.40 = £30

The total profit is 40% of this, so is ...

£30 × 40% = £12

The purchase price of the skirt is ...

total cost - glove cost = skirt cost = £30 -3 = £27

The profit on the skirt is ...

total profit - glove profit = skirt profit = £12 -100% × £3 = £9

Then the percentage profit on the skirt is ...

skirt profit % = skirt profit / skirt cost × 100% = £9/£27 × 100% = 33.3%

The percentage profit on the cost of the skirt was 33.3%.

5 0

3 years ago

PLEASE ANSWER ASAP! SHOW ALL WORK! WILL GIVE BRAINLIEST!

Iteru [2.4K]

1) $x^2=x\cdot x$ , and $x(x+3)=x^2+3x$ . Subtracting this from the numerator gives a remainder of

$(x^2-13x-48)-(x^2+3x)=-16x-48$

$-16x=-16\cdot x$ , and $-16(x+3)=-16x-48$ . Subtracting this from the previous remainder gives a new remainder of

$(-16x-48)-(-16x-48)=0$

This means that

$\dfrac{x^2-13x-48}{x+3}=x-16$

2) $3x^3=3x^2\cdot x$ , and $3x^2(x+2)=3x^3+6x^2$ . Subtracting this from the numerator gives a remainder of

$(3x^3-x^2-7x+6)-(3x^3+6x^2)=-7x^2-7x+6$

$-7x^2=-7x\cdot x$ , and $-7x(x+2)=-7x^2-14x$ . Subtracting this from the previous remainder gives a new remainder of

$(-7x^2-7x+6)-(-7x^2-14x)=7x+6$

$7x=7\cdot x$ , and $7(x+2)=7x+14$ . Subtracting this from the previous remainder gives a new remainder of

$(7x+6)-(7x+14)=-8$

This means that

$\dfrac{3x^3-x^2-7x+6}{x+2}=3x^2-7x+7-\dfrac8{x+2}$

3) $x+2$ will be a factor of $x^3+3x^2-10x-24$ if dividing the latter by $x+2$ leaves a remainder of 0.

$x^3=x^2\cdot x$ , and $x^2(x+2)=x^3+2x^2$ . Subtracting this from the numerator gives a remainder of

$(x^3+3x^2-10x-24)-(x^3+2x^2)=x^2-10x-24$

$x^2=x\cdot x$ , and $x(x+2)=x^2+2x$ . Subtracting this from the previous remainder gives a new remainder of

$(x^2-10x-24)-(x^2+2x)=-12x-24$

$-12x=-12\cdot x$ , and $-12(x+2)=-12x-24$ . Subtracting this from the previous remainder gives a new remainder of

$(-12x-24)-(-12x-24)=0$

and since the remainder is 0, $x+2$ is indeed a factor of $x^3+3x^2-10x-24$ .

5 0

3 years ago

Can you guys give me 3 examples of the inverse property

Irina-Kira [14]

Of multiplication or addition?

3 0

4 years ago

Please help me on this for brainliest and a brownie

Airida [17]

Answer:

A = 67

Step-by-step explanation:

a - 48 = 19

+48 +48

a = 67

Hope this Helps

8 0

3 years ago

Read 2 more answers

Find one value of x that is a solution to the equation (x^2+3)^2=4x^2+12

Westkost [7]

Answer:

${( {x}^{2} + 3) }^{2} = {4x}^{2} + 12$

$\bf \purple{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }$

${( {x}^{2}) }^{2} + 2({x)}^{2} (3) + {(3)}^{2} = {4x}^{2} + 12$

${x}^{4} + {6x}^{2} + 9 = {4x}^{2} + 12$

${x}^{4} = {4x}^{2} - {6x}^{2} + 12 - 9$

${x}^{4} = { - 2x}^{2} + 3$

$( {x}^{2} \times {x}^{2}) = { - 2x}^{2} + 3$

$2 \times {x}^{2} = { - 2x}^{2} + 3$

$= \frac{ { - 2x}^{2} + 3 }{ {2x}^{2} }$

$= 3$

4 0

3 years ago